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Andreas93 [3]
2 years ago
12

What is the volume of a sample if the density is 2.35 g/mL and the mass is 33.67 g?

Chemistry
1 answer:
lawyer [7]2 years ago
6 0
  • Density=2.35g/mL
  • Mass=33.67g

\\ \star\sf\longmapsto Density=\dfrac{Mass}{Volume}

\\ \star\sf\longmapsto Volume=\dfrac{Mass}{Density}

\\ \star\sf\longmapsto Volume=\dfrac{33.67}{2.35}

\\ \star\sf\longmapsto Volume=14.32mL

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Scientists repeat experiments for reliability.  Experiments have to be repeated, since performing an experiment only once, does not prove a scientists theory on the experiment, which they develop by performing the experiment.  And then performing the experiment again or even many times to prove or disapprove their theories.  Btw, before an experiment begins, the scientist will make a hypothesis of what they believe will happen.  If proven correctly, they would then use those results they record throughout the experiment, from beginning to end to prove whether or not their hypothesis are correct or incorrect.  Click to let others know, how helpful is it


Read more on Brainly.com - brainly.com/question/2613050#readmore


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7 0
3 years ago
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Which question could be answered with qualitative data?
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B

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What is the mass of 5.00 moles of KaS04?
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Answer: D

Explanation:

I assume you meant \text{K}_{2}\text{SO}_{4}.

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So, the formula mass of potassium sulfate is 2(39.0983)+32.065+4(15.9994)=174.2592 g/mol.

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7 0
2 years ago
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4.1 moles of sodium carbonate to molecules of sodium carbonate.​
docker41 [41]
<h3>Answer:</h3>

2.47 × 10^24 molecules

<h3>Explanation:</h3>

One mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23.

That is, 1 mole of a compound =  6.022 × 10^23 molecules

Therefore,

1 mole of Na₂CO₃ = 6.022 × 10^23 molecules

Thus, we can calculate the number of molecules in 4.1 moles of Na₂CO₃

we get,

 = 4.1 moles × 6.022 × 10^23 molecules

 = 2.47 × 10^24 molecules

Hence, 4.1 moles of Na₂CO₃ contains 2.47 × 10^24 molecules

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3 years ago
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